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2 years ago

In a nuclear reactor a moderator is used to

1 answer:

4 0

Answer: A material called moderator is used to slow down the neutrons released during nuclear fission. The source's neutrons have a high speed and energy. The neutrons' speed is slowed by heavy water or graphite moderators.

Explanation:

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A police radar gun uses X-band microwave radiation at a frequency of 12.2 GHz. Microwaves travel at the speed of light, or 3x108

quester [9]

Answer:

A police radar gun uses X-band microwave radiation at a frequency of 13.1 GHz. Microwaves travel at the speed of light, or 3x108 m/s. Since the frequency shift will be small for practical car speeds and difficult to detect, the shifted frequency is compared to the original frequency, and the resulting beat frequency is used to determine the speed of the car.

a.) If Michael is traveling at 29 m/s, what is the resulting beat frequency that the radar gun detects?

ANSWER: 2533 Hz

Explanation:

6 0

3 years ago

A transformer consisting of two coils wrapped around an iron core is connected to a generator and a resistor (Resistor is connec

V125BC [204]

Answer:

The peak emf of the generator is 40.94 V.

Explanation:

Given that,

Number of turns in primary coil= 11

Number of turns in secondary coil= 18

Peak voltage = 67 V

We nee to calculate the peak emf

Using relation of number of turns and emf

$\dfrac{N_{1}}{N_{2}}=\dfrac{E_{1}}{E_{2}}$

$E_{1}=\dfrac{N_{1}}{N_{2}}\times E_{2}$

Where, N₁ = Number of turns in primary coil

N₂ = Number of turns in secondary coil

E₂ = emf across secondary coil

Put the value into the formula

$E_{1}=\dfrac{11}{18}\times67$

$E_{1}=40.94\ V$

Hence, The peak emf of the generator is 40.94 V.

4 0

3 years ago

In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward

Vladimir79 [104]

Answer:

The answer is " $1.01 \times 10^{-13}$ "

Explanation:

Using the law of conservation for energy. Equating the kinetic energy to the potential energy.

$KE=U=\frac{kqq'}{r}\\\\$

Calculating the closest distance:

$\to r=\frac{kqq'}{KE}\\\\$

$=\frac{k(2e)(79e)}{KE}\\\\=\frac{k(2)(79)e^2}{KE}\\\\=\frac{9.0\times 10^9 \ N \cdot \frac{m^2}{c}(2)(79)(1.6 \times10^{-19} \ C)^2}{(2.25\ meV) (\frac{1.6 \times 10^{-13} \ J}{1 \ MeV})}\\\\$

$=\frac{9.0\times 10^9 \times 2\times 79\times 1.6 \times10^{-19}\times 1.6 \times10^{-19} }{(2.25 \times 1.6 \times 10^{-13}) }\\\\=\frac{3,640.32\times 10^{-29}}{3.6 \times 10^{-13} }\\\\=\frac{3,640.32}{3.6} \times 10^{-16}\\\\=1011.2 \times 10^{-16}\\\\=1.01 \times 10^{-13}$